The equation sqrt{x + 1} - sqrt{x - 1} = sqrt | vedclass

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(A) Given equation: $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$Squaring both sides:$(\sqrt{x + 1} - \sqrt{x - 1})^2 = (\sqrt{4x - 1})^2$$(x + 1) + (

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(A) Given equation: $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$Squaring both sides:$(\sqrt{x + 1} - \sqrt{x - 1})^2 = (\sqrt{4x - 1})^2$$(x + 1) + (x - 1) - 2\sqrt{(x + 1)(x - 1)} = 4x - 1$$2x - 2\sqrt{x^2 - 1} = 4x - 1$$-2\sqrt{x^2 - 1} = 2x - 1$Squaring both sides again:$(-2\sqrt{x^2 - 1})^2 = (2x - 1)^2$$4(x^2 - 1) = 4x^2 - 4x + 1$$4x^2 - 4 = 4x^2 - 4x + 1$$-4 = -4x + 1$$4x = 5$$x = 5/4$Check the solution in the original equation:$LHS$: $\sqrt{5/4 + 1} - \sqrt{5/4 - 1} = \sqrt{9/4} - \sqrt{1/4} = 3/2 - 1/2 = 2/2 = 1$$RHS$: $\sqrt{4(5/4) - 1} = \sqrt{5 - 1} = \sqrt{4} = 2$Since $1 
eq 2$,the value $x = 5/4$ is an extraneous solution.Therefore,the equation has no solution.

The equation $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$ has

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